Last updated at Feb. 25, 2017 by Teachoo

Transcript

Ex 9.1 , 14 A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see figure ). Find the distance travelled by the balloon during the interval. Given that 1.2 m tall girl sees a balloon So, AG = 1.2 m Also, AG & BF are parallel BF = AG = 1.2 m And the balloon is at a height of 88.2 m So, EF = 88.2 Girl sees balloon first at 60° So, ∠ EAB = 60° After travelling, the angle of elevation becomes 30° So, ∠ DAC = 30° Distance travelled by the balloon = BC We have to find BC Now, BE = EF – BF BE = 88.2 – 1.2 m BE = 87 m Also, BE = DC = 87 m Here, ∠ ABE = 90° & ∠ ACD = 90° Now, AC = AB + BC 87√3 = (" " 87)/√3 + BC 87√3 – (" " 87)/√3 = BC BC = 87√3 – (" " 87)/√3 BC = 87 (√3 – (" " 1)/√3 ) BC = 87((√3 √3 −1)/√3) BC = 87((3 − 1)/√3) BC = 87 (2/√3) BC = (" " 87 ×2)/√3 Multiply √3 in numerator and denominator BC = (" " 87 × 2)/√3 × √3/√3 BC = (" " 87 × 2 ×√3)/3 BC = 29×2×√3 BC = 58√3 Hence, distance travelled by balloon = 58√3 m

Important Questions for Exam - Class 10

- Chapter 1 Class 10 Real Numbers
- Chapter 2 Class 10 Polynomials
- Chapter 3 Class 10 Pair of Linear Equations in Two Variables
- Chapter 4 Class 10 Quadratic Equations
- Chapter 5 Class 10 Arithmetic Progressions
- Chapter 6 Class 10 Triangles
- Chapter 7 Class 10 Coordinate Geometry
- Chapter 8 Class 10 Introduction to Trignometry
- Chapter 9 Class 10 Some Applications of Trignometry
- Chapter 10 Class 10 Circles
- Chapter 11 Class 10 Constructions
- Chapter 12 Class 10 Areas related to Circles
- Chapter 13 Class 10 Surface Areas and Volumes
- Chapter 14 Class 10 Statistics
- Chapter 15 Class 10 Probability

About the Author

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He provides courses for Mathematics from Class 9 to 12. You can ask questions here.